A trigonometric of a logarithm of an imaginary power is written in a named constant times a power - #1867

Part of #718.

sec(a + 2 ln(c x^(i/2)))^3 was declined, along with x sec(a + 2 ln(c x^i))^3 and the cosecant's (Rubi's 4.7.5). The rule that writes a trigonometric function of a + b ln(u) as exponentials only fired when b itself was imaginary. Here b is 2, and the imaginary factor is the power of the variable inside the logarithm.

What the rule needs is for e^(i L) to be a constant times a power of the variable. That holds whenever b x u'/u is an imaginary constant, which for u a product of constants and powers of x is b times the sum of those powers. The condition now reads that sum off the factors.

Asked as it stands, the rewritten integrand gets right answers of over 6,000 characters, because c^(2 i) and c^(-2 i) (or e^(i a) and e^(-i a)) never cancel, so the answer is a piecewise on combinations of them. Instead, e^(i L) is written as K x^p, with K = e^(i L) x^(-p) named as one symbol. The integrand is then rational in x, and K is substituted back afterwards. K is constant wherever it is defined, for either sign of c, so the answer is an antiderivative there. Where the named answer puts a condition on K, as tan(a + i ln x)'s does on K^4 > 0, the condition has no value once K is written back. In that case the exponentials are asked unnamed, as before.

Sizes, measured on master a29f7ec, on 2.5.0 and on this branch:

Each was checked exactly, derivative against integrand at 50, 300 and 2,000 digits, for c = 2 and c = -2. All agree at every precision except one: sec(a + 2*ln(c/x^(1/2*i)))^3 at c = -2 is right at 300 and 2,000 digits but not at 50. There the integrand is about 1e-8, because ln of a negative c adds i pi, which the factor 2 i turns into e^(-2 pi). The answer's terms cancel to that, so the result is ill-conditioned rather than wrong.

The tests differentiate each one back at c = 1.3 and c = -1.3.

Rubi corpus, compared with master a29f7ec:

• 4.7.5 pocket: 145 → 150 of 160, with 0 wrong on either side.
• The 5 that moved went from unevaluated to solved, and each was re-run on its own.
• Every family is unchanged.

The allocation gate passed. The suite passed: 15,309 passed, 0 failed.

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